▶️ Ejercicio 237 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXI

R a d i c a l e s
Reducción de radicales semejantes

Ejercicio 237

Reducir:

$7 \sqrt{2} - 15 \sqrt{2} = \sqrt{2} (7 - 15) = - 8 \sqrt{2}$
$4 \sqrt{3} - 20 \sqrt{3} + 19 \sqrt{3} = \sqrt{3} (4 - 20 + 19) = 3 \sqrt{3}$
$\sqrt{5} - 22 \sqrt{5} - 8 \sqrt{5} = \sqrt{5} (1 - 22 - 8) = - 29 \sqrt{5}$
$\sqrt{2} - 9 \sqrt{2} + 30 \sqrt{2} - 40 \sqrt{2} = \sqrt{2} (1 - 9 + 30 - 40) = - 18 \sqrt{2}$
$\frac{3}{4} \sqrt{2} - \frac{1}{2} \sqrt{2} = \sqrt{2} (\frac{3}{4} - \frac{1}{2}) = \sqrt{2} (\frac{3 - 2}{4}) = \frac{1}{4} \sqrt{2}$
$\frac{3}{5} \sqrt{3} - \sqrt{3} = \sqrt{3} (\frac{3}{5} - 1) = \sqrt{3} (\frac{3 - 5}{5}) = - \frac{2}{5} \sqrt{3}$
$2 \sqrt{5} - \frac{1}{2} \sqrt{5} + \frac{3}{4} \sqrt{5} = \sqrt{5} (2 - \frac{1}{2} + \frac{3}{4}) = \sqrt{5} (\frac{8 - 2 + 3}{4}) = \frac{9}{4} \sqrt{5}$
$\frac{1}{4} \sqrt{3} + 5 \sqrt{3} - \frac{1}{8} \sqrt{3} = \sqrt{3} (\frac{1}{4} + 5 - \frac{1}{8}) = \sqrt{5} (\frac{2 + 40 - 1}{8}) = \frac{41}{8} \sqrt{5}$
$a \sqrt{b} - 3 a \sqrt{b} + 7 a \sqrt{b} = \sqrt{b} (a - 3 a + 7 a) = 5 a \sqrt{b}$
$\begin{matrix} 3 x \sqrt{y} + (a - x) \sqrt{y} - 2 x \sqrt{y} & = \sqrt{y} [3 x + (a - x) - 2 x] \\ = \sqrt{y} [3 x + a - x - 2 x] \\ = a \sqrt{y} \end{matrix}$
$\begin{matrix} (x - 1) \sqrt{3} + (x - 3) \sqrt{3} + 4 \sqrt{3} & = \sqrt{3} [(x - 1) + (x - 3) + 4] \\ = \sqrt{3} [x - 1 + x - 3 + 4] \\ = 2 x \sqrt{3} \end{matrix}$
$\frac{1}{3} \sqrt[3]{2} - \frac{2}{3} \sqrt[3]{2} + 2 \sqrt[3]{2} = \sqrt[3]{2} (\frac{1}{3} - \frac{2}{3} + 2) = \sqrt[3]{2} (\frac{1 - 2 + 6}{3}) = \frac{5}{3} \sqrt[3]{2}$
$\frac{3}{5} \sqrt[3]{2} - \frac{1}{4} \sqrt[3]{2} + \frac{1}{6} \sqrt[3]{2} = \sqrt[3]{2} (\frac{3}{5} - \frac{1}{4} + \frac{1}{6}) = \sqrt[3]{2} (\frac{36 - 15 + 10}{60}) = \frac{31}{60} \sqrt[3]{2}$
$\begin{matrix} x \sqrt[3]{a^{2}} - (a - 2 x) \sqrt[3]{a^{2}} + (2 a - 3 x) \sqrt[3]{a^{2}} & = \sqrt[3]{a^{2}} [x - (a - 2 x) + (2 a - 3 x)] \\ = \sqrt[3]{a^{2}} [x - a + 2 x + 2 a - 3 x] \\ = a \sqrt[3]{a^{2}} \end{matrix}$